The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law $$p\,V=n\,RT,$$ this comes down to $$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$ That logarithm is at fault for a lot of the formulas in chemistry.

I find I have a surprisingly hard time computing $\Delta G$ for gas governed by the equation of state $$\left(p + a\frac{n^2}{V^2}\right)\,(V - b\,n) = n\, R T,$$ where $a\ne 0,b$ are small constants. What is $\Delta G$, at least in low orders in $a,b$?

One might be able to compute ΔG by an integral in which not V is the integrand.

Edit 19.8.15: My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.

  • $\begingroup$ The change in Gibbs Free Energy for which transformative process? Going from which initial state to which final state? $\Delta G$ is a state function. $\endgroup$
    – Gert
    Aug 16 '15 at 16:42
  • $\begingroup$ @Gert: I'm not sure which aspect of your question isn't answered by the first sentence. $\endgroup$
    – Nikolaj-K
    Aug 16 '15 at 18:27

I think shifting the variable of integration should work: \begin{align} \int_{p_1}^{p_2}dp~V & =\int_\text{state 1}^\text{state 2} d(pV)-\int_{V_1}^{V_2} dV~p \\ & = p_2V_2-p_1V_1-\int_{V_1}^{V_2} dV~\left[ \frac{nRT}{V-bn}-a\frac{n^2}{V^2} \right] \end{align}


This problem contains a functional form that is very difficult to work with. A numeric integration is one approach that will give an approximate (but good) answer. To do this, the following procedure can be used:

  1. Start at pressure $P_1$, and establish a "small" value for $dP$.
  2. Use a trial-and-error method to calculate $V$.
  3. Multiply $V$ by $dP$ and keep track of the sum of $V(dP)$.
  4. Increment pressure by $dP$.
  5. Repeat steps 2-4 until $P_2$ is reached.

The size of $dP$ is arbitrary. To ensure that your value of $dP$ is appropriate, it would be helpful to calculate the integral with a value of $dP$, then divide $dP$ by $2$ and repeat the calculation. If both calculations give approximately the same answer, you are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.